Harmonic oscillators

Our proof of the equipartition theorem depends crucially on the classical approximation. To see how quantum effects modify this result, let us examine a particularly simple system which we know how to analyze using both classical and quantum physics: i.e., a simple harmonic oscillator. Consider a one-dimensional harmonic oscillator in equilibrium with a heat reservoir at temperature $T$. The energy of the oscillator is given by

$$E = \frac{p^2}{2\,m} + \frac{1}{2}\kappa\, x^2,$$ (467)

where the first term on the right-hand side is the kinetic energy, involving the momentum $p$ and mass $m$, and the second term is the potential energy, involving the displacement $x$ and the force constant $\kappa$. Each of these terms is quadratic in the respective variable. So, in the classical approximation the equipartition theorem yields:

$$\frac{\overline{p^2}}{2\,m} = \frac{1}{2}\, k\,T,$$ (468)

$$\frac{1}{2}\kappa \,\overline{x^2} = \frac{1}{2} \,k\,T.$$ (469)

That is, the mean kinetic energy of the oscillator is equal to the mean potential energy which equals $(1/2)\,k\,T$. It follows that the mean total energy is

$$\overline{E} = \frac{1}{2}\, k\,T + \frac{1}{2}\, k\,T = k\,T.$$ (470)

According to quantum mechanics, the energy levels of a harmonic oscillator are equally spaced and satisfy

$$E_n = (n + 1/2) \,\hbar \,\omega,$$ (471)

where $n$ is a non-negative integer, and

$$\omega = \sqrt{\frac{\kappa}{m}}.$$ (472)

The partition function for such an oscillator is given by

$$Z = \sum_{n=0}^\infty \exp(-\beta \,E_n) = \exp[-(1/2)\,\bet... ... \,\omega] \sum_{n=0}^\infty \exp(- n\,\beta\, \hbar\,\omega).$$ (473)

Now,

$$\sum_{n=0}^\infty \exp(- n\,\beta\, \hbar \,\omega) = 1 + \e... ...beta\,\hbar\,\omega) + \exp(-2\,\beta\,\hbar\,\omega) + \cdots$$ (474)

is simply the sum of an infinite geometric series, and can be evaluated immediately,

$$\sum_{n=0}^\infty \exp(- n\,\beta\, \hbar\, \omega) = \frac{1}{1-\exp(-\beta\,\hbar \,\omega)}.$$ (475)

Thus, the partition function takes the form

$$Z = \frac{ \exp[-(1/2)\,\beta\,\hbar\,\omega]}{1-\exp(-\beta\,\hbar\,\omega)},$$ (476)

and

$$\ln Z = - \frac{1}{2}\,\beta\,\hbar\,\omega -\ln [1- \exp(-\beta\,\hbar\,\omega)]$$ (477)

The mean energy of the oscillator is given by [see Eq. (399)]

$$\overline{E} = - \frac{\partial}{\partial \beta} \ln Z = - \... ...omega)\,\hbar\,\omega} {1-\exp(-\beta\,\hbar\,\omega)}\right],$$ (478)

or

$$\overline{E} = \hbar \,\omega \left[ \frac{1}{2} + \frac{1}{\exp(\beta\, \hbar\, \omega)-1} \right].$$ (479)

Consider the limit

$$\beta\,\hbar\,\omega = \frac{\hbar\, \omega}{k\,T} \ll 1,$$ (480)

in which the thermal energy $k\,T$ is large compared to the separation $\hbar \,\omega$ between the energy levels. In this limit,

$$\exp(\beta\,\hbar\,\omega) \simeq 1 + \beta \,\hbar\, \omega,$$ (481)

so

$$\overline{E} \simeq \hbar\,\omega\left[\frac{1}{2} + \frac{1... ...meq \hbar\,\omega\left[ \frac{1}{\beta\,\hbar\,\omega}\right],$$ (482)

giving

$$\overline{E} \simeq \frac{1}{\beta} = k\,T.$$ (483)

Thus, the classical result (470) holds whenever the thermal energy greatly exceeds the typical spacing between quantum energy levels.

Consider the limit

$$\beta\,\hbar\,\omega = \frac{\hbar \,\omega}{k\,T} \gg 1,$$ (484)

in which the thermal energy is small compared to the separation between the energy levels. In this limit,

$$\exp(\beta\,\hbar\,\omega) \gg 1,$$ (485)

and so

$$\overline{E} \simeq \hbar\,\omega \,[ 1/2 + \exp(-\beta\,\hbar\,\omega)] \simeq \frac{1}{2} \,\hbar \,\omega.$$ (486)

Thus, if the thermal energy is much less than the spacing between quantum states then the mean energy approaches that of the ground-state (the so-called zero point energy). Clearly, the equipartition theorem is only valid in the former limit, where $k\,T \gg \hbar\, \omega$, and the oscillator possess sufficient thermal energy to explore many of its possible quantum states.